Lecture 4 CS781 February 3, 2011

Transcription

1 Lecture 4 CS78 February 3, 2 Topics: Data Compression-Huffman Trees Divide-and-Conquer Solving Recurrence Relations Counting Inversions Closest Pair Integer Multiplication Matrix Multiplication

2 Data Compression and Huffman Codes

3 The Problem of Encoding Encoding Symbols using bit-strings. Before storing or transmitting with computers, we tend to construct files as logical sequences of symbols. The problem of data compression is to take a file of symbols and encode it using a bit-string so that it can be effectively reconstructed later in time or in another place. Variable-length code. ASCII codes are used to encode keyboard symbols using a fixed number (7) bits per symbol. This may not be the most efficient encoding since some symbols are used much more frequently than others, and therefore should be encoded using fewer bits. Prefix Codes. Variable-length codes known as prefix codes address this problem. Prefix codes have the property that each codeword has a unique prefix this is an important property that allows the unambiguous reconstruction of the file. 3

4 Optimal Prefix Codes The Problem of Optimum Prefix Codes. Given an set of symbols S and a file F containing a sequence of symbols to encode. We want to find a prefix code (one codeword per symbol) which minimizes the size of the encoded file. Let frequency f(x) = how many times symbol x appears in file. Let bin(x) denote the binary code of a symbol x. The size of the encoded file is therefore the sum x f(x) length(bin(x)) Coding Tree. A prefix code is presented by a binary coding tree where each leaf stores, and represents one symbol. Each internal node of the tree has two children (left ~ ; right ~ ). A codeword for a symbol is given by the unique path from the root to the leaf. 4

5 Huffman Coding Algorithm The depth of a leaf w(x) of the coding tree is the distance from the root to the leaf. Given a coding tree T, the cost of of the tree is size of the encoded file cost(t) = x f(x) depth(w(x)) Huffman Coding Algorithm builds a tree of optimal cost. Input: List [f,f2,,fn] of frequencies of n symbols [s, s2,, sn] Output: Optimal prefix coding tree, which minimizes total cost over all coding trees.

6 Huffman Coding Algorithm Huffman Coding Algorithm Input: List [f,f2,,fn] of frequencies of n symbols [s, s2,, sn] Output: Optimal prefix coding tree, which minimizes total cost over all coding trees.. Sort the symbols by frequency f < f2 <.. < fn 2. Create n leaf nodes one for each of the n symbols (s,f), (s2,f2),, (sn,fn) 3. Take the two least frequent symbols (si,fi), (sj,fj) in the alphabet and link these two symbols into a single parent node with combined frequencies, (sij, fi+ fj) 4. Repeat step 3, until the tree is finished (n- iterations).

7 Example of Huffman Tree si=[ a, b, c, d, e ] fi=[.25,.25,.2,.5,.5] a.25 b.25 c.2 d.5 e.5

8 Optimality of the Huffman Code Proof Sketch (by induction) Note that the algorithm is recursive, it invokes the same process on smaller sets of symbols. So lets use induction. If there are just one or two symbols the algorithm is optimal as it uses just bit per symbol. Suppose we assume that it is optimal for all k-symbol prefix codes. Given a k+ symbol problem, recall that the algorithm builds a tree T by first selecting the two lowest frequency symbols say x,y and merges them. By induction we can assume that the algorithm returns an optimal encoding tree for this transformed file of symbols (x,y identified). Suppose T was not of optimal cost, there is a optimal tree T such that cost(t ) < cost(t). We can argue that T must have x,y at lowest levels and that they are siblings in T, since it is optimal. Merging these symbols in T will result in a tree that has lower cost than optimal for the k-symbol transformed problem. QED.

9 Implementation and Analysis of Huffman Algorithm Analysis of Huffman Algorithm. There are clearly n- iterations to solve an n-symbol problem. In each iteration we need to select the two lowest frequency symbols. Using a priority queue, we can extract-min in time O(log n). We then merge into a new parent node with key value equal to the sum of the two lowest frequency symbols. This node is inserted into the priority queue in O(log n) time. Summing over all iterations we have complexity of O(n log n). Python Implementation of Huffman Algorithm import heapq - Priority Queue module using binary heaps heapq.heapify(x) - Transforms list x into a heap, in-place, in linear time. heapq.heappush(heap, item) Push item onto the heap, maintaining heap invariant. heapq.heappop(heap) - Pop and return the smallest item from the heap, maintaining the heap invariant.

10 import heapq def makehufftree(symboltuplelist): Python Code for Huffman trees = list(symboltuplelist) Tree heapq.heapify(trees) while len(trees) > : childr, childl = heapq.heappop(trees), heapq.heappop(trees) parent = (childl[] + childr[], childl, childr) heapq.heappush(trees, parent) return trees[] def printhufftable(hufftree, prefix = ''): global totallen if len(hufftree) == 2: print hufftree[],hufftree[], prefix totallen=totallen+(hufftree[]*len(prefix)) else: printhufftable(hufftree[], prefix + '') printhufftable(hufftree[2], prefix + '') cincinnati=[(2,'c'),(3,'i'),(3,'n'),(,'a'),(,'t')] hufftree = makehufftree(cincinnati) global totallen; totallen = printhufftable(hufftree) print 'total length is', totallen >>> 3 n 3 i t a 2 c total length is 22 >>>

11 Disadvantages of the Huffman Code Changing frequencies If the frequencies and probabilities change the optimal coding changes e.g. in text compression symbol frequencies vary with context Re-computing the Huffman code by running through the entire file in advance?! Saving/ transmitting the code too?! Does not consider blocks of symbols strings_of_ch the next nine symbols are predictable aracters_, but bits are used without conveying any new information

12 Divide-and-Conquer Divide-and-conquer. Break up problem into several parts. Solve each part recursively. Combine solutions to sub-problems into overall solution. Most common usage. Break up problem of size n into two equal parts of size ½n. Solve two parts recursively. Combine two solutions into overall solution in linear time. Consequence. Brute force: n 2. Divide-and-conquer: n log n. 2

13 Mergesort Mergesort. Divide array into two halves. Recursively sort each half. Merge two halves to make sorted whole. Jon von Neumann (945) A L G O R I T H M S A L G O R I T H M S divide O() A G L O R H I M S T sort 2T(n/2) A G H I L M O R S T merge O(n) 3

14 A Useful Recurrence Relation Def. T(n) = number of comparisons to mergesort an input of size n. Mergesort recurrence. T(n) if n = ( ) T n/2 solve left half ( ) + T n/2 solve right half + n merging otherwise Solution. T(n) = O(n log 2 n). Assorted proofs. We describe several ways to prove this recurrence. Initially we assume n is a power of 2 and replace with =. 4

15 Proof by Recursion Tree T(n) = if n = 2T(n/2) sorting both halves + n merging otherwise T(n) n T(n/2) T(n/2) 2(n/2) T(n/4) T(n/4) T(n/4) T(n/4) log 2 n 4(n/4)... T(n / 2 k ) 2 k (n / 2 k )... T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) n/2 (2) n log 2 n 5

16 Proof by Telescoping or Unrolling Claim. If T(n) satisfies this recurrence, then T(n) = n log 2 n. assumes n is a power of 2 T(n) = if n = 2T(n/2) sorting both halves + n merging otherwise Pf. For n > : T(n) n = 2T(n /2) n + = T(n /2) n /2 + = = T(n / 4) n / 4 T(n /n) n /n = log 2 n log 2 n 6

17 Proof by Induction Claim. If T(n) satisfies this recurrence, then T(n) = n log 2 n. T(n) = Pf. (by induction on n) Base case: n =. if n = 2T(n/2) sorting both halves + n merging Inductive hypothesis: T(n) = n log 2 n. Goal: show that T(2n) = 2n log 2 (2n). otherwise assumes n is a power of 2 T(2n) = 2T(n) + 2n = 2nlog 2 n + 2n ( ) + 2n = 2n log 2 (2n) = 2nlog 2 (2n) 7

18 Analysis of Mergesort Recurrence Claim. If T(n) satisfies the following recurrence, then T(n) n lg n. T(n) if n = ( ) T n/2 solve left half ( ) + T n /2 + n solve right half merging otherwise log 2 n Pf. (by induction on n) Base case: n =. Define n = n / 2, n 2 = n / 2. Induction step: assume true for, 2,..., n. T(n) T(n ) + T(n 2 ) + n n lgn + n 2 lg n 2 + n n lgn 2 + n 2 lg n 2 + n = n lgn 2 + n n( lg n ) + n = n lgn n 2 = n/2 2 lg n / 2 = 2 lg n / 2 lgn 2 lg n 8

19 Counting Inversions

20 Counting Inversions Music site tries to match your song preferences with others. You rank n songs. Music site consults database to find people with similar tastes. Similarity metric: number of inversions between two rankings. My rank:, 2,, n. Your rank: a, a 2,, a n. Songs i and j inverted if i < j, but a i > a j. Songs Me You A B C D E Inversions 3-2, 4-2 Brute force: check all Θ(n 2 ) pairs i and j. 2

21 Counting Inversions: Divide-and-Conquer Divide-and-conquer

22 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate list into two pieces Divide: O()

23 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate list into two pieces. Conquer: recursively count inversions in each half Divide: O() Conquer: 2T(n / 2) 5 blue-blue inversions 8 green-green inversions 5-4, 5-2, 4-2, 8-2, , 9-3, 9-7, 2-3, 2-7, 2-, -3, -7 23

24 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate list into two pieces. Conquer: recursively count inversions in each half. Combine: count inversions where a i and a j are in different halves, and return sum of three quantities Divide: O() Conquer: 2T(n / 2) 5 blue-blue inversions 8 green-green inversions 9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, -6, -9, -3, -7 Combine:??? Total = =

25 Counting Inversions: Combine Combine: count blue-green inversions Assume each half is sorted. Count inversions where a i and a j are in different halves. Merge two sorted halves into sorted whole. to maintain sorted invariant blue-green inversions: Count: O(n) Merge: O(n) T(n) T n/2 ( ) + T n/2 ( ) + O(n) T(n) = O(nlog n) 25

26 DeMerge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 26

27 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves 6 2 auxiliary array Total: 6 27

28 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves 6 2 auxiliary array Total: 6 28

29 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 29

30 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 3

31 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 3

32 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 32

33 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 33

34 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 34

35 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

36 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

37 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

38 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

39 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

40 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

41 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

42 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

43 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

44 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

45 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

46 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. first half exhausted i = two sorted halves auxiliary array Total:

47 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

48 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

49 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

50 Merge and Count Merge and count step. Given two sorted halves, count number of inversions where a i and a j are in different halves. Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: = 3 5

51 Counting Inversions: Implementation Pre-condition. [Merge-and-Count] A and B are sorted. Post-condition. [Sort-and-Count] L is sorted. >>> Sort_and_Count([6,5,3,,4,2,]) (7, [, 2, 3, 4, 5, 6, ]) def Sort_and_Count(L): if len(l)== : return (,L) A = L[:len(L)/2] B = L[len(L)/2:] (inva, A) = Sort_and_Count(A) (invb, B)= Sort_and_Count(B) (invab, L) = Merge_and_Count(A, B) inv = inva + invb + invab return (inv, L) def Merge_and_Count(A, B): count=; result = [] i,j =, while i < len(a) and j < len(b): if A[i] <= B[j]: result.append(a[i]) i += else: result.append(b[j]) j += count += len(a) - i result += A[i:] + B[j:] return (count,result) 5

52 Closest Pair of Points

53 Closest Pair of Points Closest pair. Given n points in the plane, find a pair with smallest Euclidean distance between them. Fundamental geometric primitive. Graphics, computer vision, geographic information systems, molecular modeling, air traffic control. Brute force. Check all pairs of points p and q with Θ(n 2 ) comparisons. -D version. O(n log n) easy if points are on a line. How would you divide-and-conquer in -D. Simplifying Assumption. No two points have same x coordinate. 53

54 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. L 54

55 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. Obstacle. Impossible to ensure n/4 points in each piece. L 55

56 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. L 56

57 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. Conquer: find closest pair in each side recursively. L

58 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. Conquer: find closest pair in each side recursively. Combine: find closest pair with one point in each side. Return best of 3 solutions. seems like Θ(n 2 ) L

59 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < δ. L 2 2 δ = min(2, 2) 59

60 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < δ. Observation: only need to consider points within δ of line L. L 2 2 δ = min(2, 2) δ 6

61 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < δ. Observation: only need to consider points within δ of line L. Sort points in 2δ-strip by their y coordinate. 7 L δ = min(2, 2) 2 δ 6

62 Closest Pair of Points Find closest pair with one point in each side, assuming that distance < δ. Observation: only need to consider points within δ of line L. Sort points in 2δ-strip by their y coordinate. Only check distances of those within positions in sorted list! 7 L δ = min(2, 2) 2 δ 62

63 Closest Pair of Points Def. Let s i be the point in the 2δ-strip, with the i th smallest y-coordinate. Claim. If i j 2, then the distance between s i and s j is at least δ. Pf j No two points lie in same ½δ-by-½δ box. Two points at least 2 rows apart have distance 2(½δ). 2 rows 29 3 ½δ ½δ Fact. Still true if we replace 2 with 7. i ½δ δ δ 63

64 Closest Pair Algorithm Closest-Pair(p,, p n ) { Compute separation line L such that half the points are on one side and half on the other side. δ = Closest-Pair(left half) δ 2 = Closest-Pair(right half) δ = min(δ, δ 2 ) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and compare distance between each point and next neighbors. If any of these distances is less than δ, update δ. O(n log n) 2T(n / 2) O(n) O(n log n) O(n) } return δ. 64

65 Closest Pair of Points: Analysis Running time. T(n) 2T( n/2) + O(n log n) T(n) = O(n log 2 n) Q. Can we achieve O(n log n)? A. Yes. Don't sort points in strip from scratch each time. Each recursive call returns two lists: all points sorted by y coordinate, and all points sorted by x coordinate. Sort by merging two pre-sorted lists. T(n) 2T( n/2) + O(n) T(n) = O(n log n) 65

66 Integer Multiplication

67 67 Integer Arithmetic Add. Given two n-digit integers a and b, compute a + b. O(n) bit operations. Multiply. Given two n-digit integers a and b, compute a b. Brute force solution: Θ(n 2 ) bit operations. * + Add Multiply

68 Divide-and-Conquer Multiplication: Warmup To multiply two n-digit integers: Multiply four ½n-digit integers. Add two ½n-digit integers, and shift to obtain result. x = 2 n /2 x + x y = 2 n /2 y + y ( ) ( 2 n /2 y + y ) = 2 n x y + 2 n /2 x y + x y xy = 2 n /2 x + x ( ) + x y ( ) T(n) = 4T n /2 + Θ(n) T(n) = Θ(n2 ) recursive calls add, shift assumes n is a power of 2 68

69 Karatsuba Multiplication To multiply two n-digit integers: Add two ½n digit integers. Multiply three ½n-digit integers. Add, subtract, and shift ½n-digit integers to obtain result. Anatoli Karatsuba x = 2 n /2 x + x y = 2 n /2 y + y xy = 2 n x y + 2 n /2 x y + x y Theorem. [Karatsuba-Ofman, 962] Can multiply two n-digit integers in O(n.585 ) bit operations. ( ) + x y ( ) + x y = 2 n x y + 2 n /2 (x + x )(y + y ) x y x y A B A C C ( ) + T n /2 ( ) + T( + n /2 ) T(n) T n/2 recursive calls T(n) = O(n log 2 3 ) = O(n.585 ) + Θ(n) add, subtract, shift 69

70 Karatsuba: Recursion Tree T(n) = if n = 3T(n/2) + n otherwise log 2 n T(n) = n 2 3 = k= ( ) k ( ) +log 2 n = 3n log T(n) n T(n/2) T(n/2) T(n/2) 3(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) 9(n/4) T(n / 2 k ) 3 k (n / 2 k ) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) 3 lg n (2) 7

71 Matrix Multiplication

72 Matrix Multiplication Matrix multiplication. Given two n-by-n matrices A and B, compute C = AB. n c ij = a ik b kj k= c c 2 c n c 2 c 22 c 2n c n c n2 c nn = a a 2 a n a 2 a 22 a 2n a n a n2 a nn b b 2 b n b 2 b 22 b 2n b n b n2 b nn Brute force. Θ(n 3 ) arithmetic operations. Fundamental question. Can we improve upon brute force? 72

73 Matrix Multiplication: Warmup Divide-and-conquer. Divide: partition A and B into ½n-by-½n blocks. Conquer: multiply 8 ½n-by-½n recursively. Combine: add appropriate products using 4 matrix additions. C C 2 = A A 2 B B 2 C 2 C 22 A 2 A 22 B 2 B 22 ( ) + ( A 2 B 2 ) ( ) + ( A 2 B 22 ) ( ) + ( A 22 B 2 ) ( ) + ( A 22 B 22 ) C = A B C 2 = A B 2 C 2 = A 2 B C 22 = A 2 B 2 ( ) T(n) = 8T n/2 + Θ(n2 ) recursive calls add, form submatrices T(n) = Θ(n 3 ) 73

74 Matrix Multiplication: Key Idea Key idea. multiply 2-by-2 block matrices with only 7 multiplications. C C 2 = C 2 C 22 A A 2 B B 2 A 2 A 22 B 2 B 22 P = A (B 2 B 22 ) P 2 = ( A + A 2 ) B 22 C = P 5 + P 4 P 2 + P 6 C 2 = P + P 2 C 2 = P 3 + P 4 C 22 = P 5 + P P 3 P 7 P 3 = ( A 2 + A 22 ) B P 4 = A 22 (B 2 B ) P 5 = ( A + A 22 ) (B + B 22 ) P 6 = ( A 2 A 22 ) (B 2 + B 22 ) P 7 = ( A A 2 ) (B + B 2 ) 7 multiplications. 8 = + 8 additions (or subtractions). 74

75 Fast Matrix Multiplication Fast matrix multiplication. (Strassen, 969) Divide: partition A and B into ½n-by-½n blocks. Compute: 4 ½n-by-½n matrices via matrix additions. Conquer: multiply 7 ½n-by-½n matrices recursively. Combine: 7 products into 4 terms using 8 matrix additions. Analysis. Assume n is a power of 2. T(n) = # arithmetic operations. ( ) T(n) = 7T n /2 + Θ(n2 ) recursive calls add, subtract T(n) = Θ(n log 2 7 ) = O(n 2.8 ) 75

76 Fast Matrix Multiplication in Practice Implementation issues. Sparsity. Caching effects. Numerical stability. Odd matrix dimensions. Crossover to classical algorithm around n = 28. Common misperception: "Strassen is only a theoretical curiosity." Advanced Computation Group at Apple Computer reports 8x speedup on G4 Velocity Engine when n ~ 2,5. Range of instances where it's useful is a subject of controversy. Remark. Can "Strassenize" Ax=b, determinant, eigenvalues, and other matrix ops. 76

77 Fast Matrix Multiplication in Theory Q. Multiply two 2-by-2 matrices with only 7 scalar multiplications? A. Yes! [Strassen, 969] Θ(n log 2 7 ) = O(n 2.8 ) Q. Multiply two 2-by-2 matrices with only 6 scalar multiplications? A. Impossible. [Hopcroft and Kerr, 97] Θ(n log 2 6 ) = O(n 2.59 ) Q. Two 3-by-3 matrices with only 2 scalar multiplications? A. Also impossible. Θ(n log 3 2 ) = O(n 2.77 ) Q. Two 7-by-7 matrices with only 43,64 scalar multiplications? A. Yes! [Pan, 98] Θ(n log ) = O(n 2.8 ) Decimal wars. December, 979: O(n ). January, 98: O(n ). 77

78 Fast Matrix Multiplication in Theory Best known. O(n ) [Coppersmith-Winograd, 987.] Conjecture. O(n 2+ε ) for any ε >. Caveat. Theoretical improvements to Strassen are progressively less practical. 78